Question

M, a solid cylinder (M=2.11 kg, R=0.137 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.870 kg mass, i.e., F = 8.535 N.
A. Calculate the angular acceleration of the cylinder.

For A. I got this and it's correct "59.0 rad/s^2"


B. If instead of the force F an actual mass m = 0.870 kg is hung from the string, find the angular acceleration of the cylinder.



C.How far does m travel downward between 0.670 s and 0.870 s after the motion begins

Answer 1

Here is the Image of the figure!

Answer 2

The force is equal to the weight of a hanging mass. This doesnt mean that there is actually a mass hanging off of it, it means there is a constant force applied of 8.142N.
Once you see that the problem is simple.
You have the equation T=I*alpha
alpha is angular acceleration which is what you need so just solve for alpha.
alpha=T/I where T is the torque applied and I is the moment of inertia.
T = F*r (when torque is perpendicular)
T = (8.142N) (0.133m)
T= 1.083Nm

Now find I, since it is a solid cylinder
I=(1/2)Mr^2
I = (1/2) (1.99kg) (0.133m)^2
I = 0.0176 kg*m^2

now recall
alpha = T/I
alpha = (1.083Nm)/(.0176kg*m^2)

ALPHA = 61.5 rad/s^2

Answer 3

@Alonzo19